➕ Lesson 6: Vector Components & Addition

Breaking vectors into components — the key to 2D forces, motion, and fields

🎯 Learning Objectives

• Resolve any vector into x- and y-components
• Add vectors by components (analytical method)
• Find magnitude and direction of resultant
• Apply to forces, velocity, acceleration, momentum, fields

🔑 Key Concepts

Vector Components

A vector A with magnitude A and angle θ from +x axis:

Aˣ = A cos θ     Aʸ = A sin θ

Conversely: A = √(Aˣ² + Aʸ²), θ = tan⁻¹(Aʸ/Aˣ)

Adding Vectors by Components

1. Resolve each vector into x and y components
2. Add all x-components → Rˣ
3. Add all y-components → Rʸ
4. Resultant: R = √(Rˣ² + Rʸ²), θ = tan⁻¹(Rʸ/Rˣ)

Unit Vector Notation

A = Aˣî + Aʸĵ (î = unit vector in x, ĵ = in y)

This makes addition trivial: just add î components and ĵ components separately.

Physics Applications

Quantity	Symbol	Component Form
Displacement	Δr	Δx î + Δy ĵ
Velocity	v	vₓ î + vʸ ĵ
Acceleration	a	aₓ î + aʸ ĵ
Force	F	Fₓ î + Fʸ ĵ
Momentum	p	pₓ î + pʸ ĵ
Electric Field	E	Eₓ î + Eʸ ĵ

📐 Worked Examples

Example 1: Force Components

Three forces act on an object: F₁ = 10 N at 0°, F₂ = 15 N at 90°, F₃ = 20 N at 210°. Find net force.

F₁ˣ = 10 cos0° = 10,    F₁ʸ = 10 sin0° = 0
F₂ˣ = 15 cos90° = 0,    F₂ʸ = 15 sin90° = 15
F₃ˣ = 20 cos210° = 20(-√3/2) = -17.3,    F₃ʸ = 20 sin210° = 20(-½) = -10

Rˣ = 10 + 0 - 17.3 = -7.3 N
Rʸ = 0 + 15 - 10 = 5 N
R = √((-7.3)² + 5²) = √(53.3+25) = √78.3 = 8.85 N
θ = tan⁻¹(5/-7.3) = tan⁻¹(-0.685) = -34.4° + 180° = 145.6° (QII)
            

Example 2: Projectile Velocity Components

Projectile launched at 30 m/s, 40°. Find vₓ, vᵧ. After 2s, find velocity components (g=10).

v₀ₓ = 30 cos40° = 23.0 m/s
v₀ᵧ = 30 sin40° = 19.3 m/s

vₓ(2s) = 23.0 m/s (constant, no air resistance)
vᵧ(2s) = v₀ᵧ - gt = 19.3 - 20 = -0.7 m/s (descending)
Speed = √(23.0² + 0.7²) = 23.0 m/s
Angle = tan⁻¹(-0.7/23.0) = -1.7° (slightly below horizontal)
            

Example 3: Incline Plane — Component Method

Block on 30° incline, m=5kg, μ=0.2. Find acceleration using vector components.

Choose axes: x down incline, y perpendicular to incline.
Weight components: Wₓ = mg sin30° = 50 × 0.5 = 25 N
                   Wᵧ = -mg cos30° = -50 × (√3/2) = -43.3 N
Normal force: Nᵧ = +43.3 N (cancels Wᵧ)
Friction: f = μN = 0.2 × 43.3 = 8.66 N (opposes motion, so -x direction)

Net Fₓ = 25 - 8.66 = 16.34 N
aₓ = Fₓ/m = 16.34/5 = 3.27 m/s²
            

🏋️ Practice Problems

1. Resolve into components: (a) 50 N at 30° (b) 100 m/s at 60° (c) 200 N at 225°
2. Find resultant: A = 20î + 10ĵ, B = -15î + 25ĵ
3. Three forces: 10N at 0°, 10N at 120°, 10N at 240°. Net force? (Symmetry!)
4. Projectile: 25 m/s at 50°. v components at t=0, t=1s, t=3s (g=10).
5. Two forces: F₁=30N @ 30°, F₂=40N @ 150°. Find resultant magnitude/direction.
6. Vector A = 5î - 12ĵ. Find magnitude, angle (in standard position).
7. Relative velocity: Boat 5 m/s north, current 3 m/s east. Velocity relative to ground?

✅ Answers

Click to reveal answers

1) (a) 43.3î+25ĵ (b) 50î+86.6ĵ (c) -141î-141ĵ 2) R=5î+35ĵ, R=35.4, θ=81.9° 3) Zero (symmetric!)

4) v₀=(16.1,19.1); t=1: (16.1,9.1); t=3: (16.1,-10.9) 5) R=36.1N, θ=112.6° 6) 13, -67.4° (or 292.6°) 7) √34≈5.83 m/s, tan⁻¹(5/3)≈59° N of E

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